= Solution
In the ionic assignment, both tetrahedral anions would require Mn(VII), $d^0$. Such a high <oxidation state> is strongly oxidizing and is best supported by strong bonding to <oxygen>. Short Mn–O bonds combine good <orbital overlap> with <oxygen>'s resistance to further oxidation. This stabilizes <permanganate> relative to a hypothetical all-<sulfide> analogue.
<sulfide> is larger, more polarizable and much easier to oxidize than <oxide>. An attempted Mn(VII) <sulfide> would therefore combine an exceptionally oxidizing <metal> center with reducing ligands: internal <redox> can reduce <manganese> while oxidizing <sulfur>, for example toward S–S-bonded or elemental <sulfur> products. The larger <sulfur> donors also provide less favorable short multiple bonding to the small, highly charged <manganese> center. \b[The <oxygen> compound supports Mn(VII); the <sulfide> combination is <redox>-incompatible.] This predicts instability without requiring a specific unprovided decomposition product.
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